$f(x) = \dfrac{ \sqrt{ x - 2 } }{ x^2 + 13 x + 40 }$ What is the domain of the real-valued function $f(x)$ ?
Solution: $f(x) = \dfrac{ \sqrt{ x - 2 } }{ x^2 + 13 x + 40 } = \dfrac{ \sqrt{ x - 2 } }{ ( x + 5 )( x + 8 ) }$ First, we need to consider that $f(x)$ is undefined anywhere where the radical is undefined, so the radicand (the expression under the radical) cannot be less than zero. So $x - 2 \geq 0$ , which means $x \geq 2$ Next, we also need to consider that $f(x)$ is undefined anywhere where the denominator is zero. So $x \neq -5$ and $x \neq -8$ However, these last two restrictions are irrelevant since $2 > -5$ and $2 > -8$ and so $x \geq 2$ will ensure that $x \neq -5$ and $x \neq -8$ Combining these restrictions, then, leaves us with simply $x \geq 2$ Expressing this mathematically, the domain is $\{ \, x \in \RR \mid x \geq2\, \}$.